385 Mini Parser

385. Mini Parser

1. Question

Given a nested list of integers represented as a string, implement a parser to deserialize it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Note: You may assume that the string is well-formed:

• String is non-empty.

• String does not contain white spaces.

• String contains only digits0-9,[,-,,].

Example 1:

Given s = "324",

You should return a NestedInteger object which contains a single integer 324.

Example 2:

Given s = "[123,[456,[789]]]",

Return a NestedInteger object containing a nested list with 2 elements:

1. An integer containing value 123.
2. A nested list containing two elements:
    i.  An integer containing value 456.
    ii. A nested list with one element:
     a. An integer containing value 789.

2. Implementation

(1) Stack

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *     // Constructor initializes an empty nested list.
 *     public NestedInteger();
 *
 *     // Constructor initializes a single integer.
 *     public NestedInteger(int value);
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // Set this NestedInteger to hold a single integer.
 *     public void setInteger(int value);
 *
 *     // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 *     public void add(NestedInteger ni);
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
class Solution {
    public NestedInteger deserialize(String s) {
        if (s.charAt(0) != '[') {
            return new NestedInteger(Integer.parseInt(s));
        }

        Stack<NestedInteger> stack = new Stack<>();

        int n = s.length();

        NestedInteger res = null;
        NestedInteger curNI = new NestedInteger();
        int i = 0;
        while (i < n) {
            char c = s.charAt(i);
            if (c == '[') {
                curNI = new NestedInteger();
                if (!stack.isEmpty()) {
                    stack.peek().add(curNI);
                }
                stack.push(curNI);
            }
            else if (c == ']') {
                res = stack.pop();
            }
            else if (c == '-' || Character.isDigit(c)) {
                int sign = 1;
                int val = 0;
                if (c == '-') {
                    sign = -1;
                }
                else {
                    val = c - '0';
                }
                // skip current position
                ++i;

                while (i < n && Character.isDigit(s.charAt(i))) {
                    val = 10 * val + s.charAt(i) - '0';
                    ++i;
                }
                // return back last position, because we will increase i in the end
                --i;

                curNI = new NestedInteger(sign * val);
                if (!stack.isEmpty()) {
                    stack.peek().add(curNI);
                }
            }
            ++i;
        }
        return res;
    }
}

3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(n)